Sunday, November 14, 2010

This weekend, I attended a conference about geometry and topology. Last night was the conference banquet and there I was sitting in a table at dinner time with a bunch of mathematicians. After some egg rolls and a wonton soup, and obvious topic made its appearance: prime numbers.

My friend Brian Streit was sitting next to me, and in the middle of the dinner he proposed this interesting and rather old question:

Given any string of digits, can one find a prime number ending in such a string? How about beginning in it?

My immediate reaction was to answer yes to both questions, as I recall seen them before in my early olympic training days.

That also made me remember a conjecture that was given to me few months ago by one of my friends back home, Rafa Martinez, who was claiming the following:

Given any string of digits, there exists a digit such that you can insert it in at a position in the original string such that the resulting number is prime.

This statement is in a sense much stronger than the previous one, and as strong as it seems, it turns out to be false, but thinking about it can make you discover some interesting properties about prime numbers and base representation of numbers.

When Rafa gave me this problem, he was so convinced of its validity and I was so convinced of its falsity, that I started looking for a counterexample. First, I started with a single digit prime number, then I start looking for two digit prime numbers that contained the first one. Then I looked for 3-digit prime numbers that contained the previous one and so on.

The first counterexample that I found by this method was 612113, but thinking of this led me to state the same question but in a different base, say $b$. Of course, the statement will still be false, but now, depending on $b$, the length of the counterexample would be different. For instance, if we are in base 2, a counterexample would be 1 (which is not prime), in base 3 we have for example 212, etc.

Basically, this phenomenon is due to the rate of growth of the prime numbers and how that can be related with the base in which numbers are written in.

Broadly speaking, we can make use of the prime number theorem to establish the rate of growth of prime numbers. It basically says, in simple words, that prime numbers grow in a logarithmic way, notice I'm not talking about the density of the primes, but about the size of a prime itself.

Hence, there is a high probability that if a number $p$ is prime, then "$ep$" is also a prime number, which, in some sense says that the most natural "base" to write prime numbers in is un "base $e$".

Of course $ep$ is not a prime number, as it is irrational, even more, what does "base $e$" mean anyway? The idea is to use this approach to find some appropriate bases for writing prime numbers. An appropriate base would be one such $b$ is close to a power of $e$. The first of these such powers is

Of course, changing base representation wont make the behavior of prime numbers different or anything like that, but maybe using this representations could lead to nice patterns on its representations.

After this little digression, I was then left with the original question of Brian, which I didn't think until today in the morning when I was discussing it about with other participant of the conference.

About the question that if you could find a prime number starting with a given string of digits (not ending in even or 5,0 of course) the simplest solution I could think was using Dirichlet's theorem on arithmetic progressions. Given a string of digits $a$, we look at the sequence

$a+k10^n$

with $k$ running over the positive integers and $n$ been the length of $a$. Hence Dirichlet's theorem asserts that there are in fact, infinitely many primes in this sequence, and hence, infinitely many numbers answering positively the question posed.

For the remaining question, my proof didn't come as quickly as the previous one, and it took me the entire last talk (which sadly I didn't pay attention to) to come up with a proof, aided with my good old friend Wikipedia.

I found this interesting fact about prime gaps due to Hoheisel, that states that

$p_{n+1}-p_n \leq p_n^\theta$ for $\theta$ less than 1

where $p_n$ is the $n$th prime number.

With the help of this, proving the statement is not really hard. Proceeding by contradiction, suppose that $a$ is a given string of digits, and that there is no prime number in between $a10^m$ and $(a+1)10^m$ for all $m$, hence this would imply that the relative gap of prime numbers would be of at least

$\frac{ a 10^m- (a+1) 10^m}{a 10^m}=\frac{1}{a} $

but by Hoheisel's result, as $n\to\infty$

$ \frac{p_{n+1}-p_n}{p_n}\to 0$

which leads to a contradiction.

I particularly find really interesting this connections between prime distributions and base representations, it seems that we don't completely understand yet the interaction of these two, but definitely there is an intimate relation among them.

Tuesday, July 13, 2010

Few days ago, I went to a summer school in algebraic geometry in Peru, and ithere was one course in projectivization of some special types of curves and properties of them.

One of the examples was doing the projectivization of a general conic over $\mathbb{C}^n$ and finding some nice properties that the resulting curves have.

The idea of the projectivization of a curve, is to grab the domain of definition and then do its compactification, for instance, in the case of a conic in $\mathbb{C}^2$, the result is to have a conic in a sphere (Riemann sphere). Algebraically, the idea is to have a curve defined as the zero set of a polynomial, for example, $p(x,y)=ax^2+bxy+cy^2=0$ with $(a,b,c)\in\mathbb{C}^3$, and then to extend the domain where the parameters are defined, i.e. $[a:b:c]\in\mathbb{P}(\mathbb{C}^2)$.

The speaker was talking about the cases in dimension 2 and 3, where the proyective spaces are $\mathbb{P}(\mathbb{C}^2)$ and $\mathbb{P}(\mathbb{C}^5)$ , and she defined a nice realization of the later one as the set of $3\times 3$ complex symmetric matrices, i.e.

This reminded me of the good old $O(3)$ with some differences. First of all, we have complex entries in our matrix, and second, the entries must satisfy and extra condition, we identify set of parameters with a common factor ($A\sim B$ iff $A=\lambda B$ for some $\lambda\in\mathbb{C}^x$), that is, the matrix defines the same curve up to a scalar multiple, which means that we need to scale it in some sense.

This two things can be fixed by considering $SO(3)$ instead of the whole $O(3)$ and then, by doing its complexification, $SO(3)\times_{\mathbb{R}}\mathbb{C}$.

So, my natural conjecture was that

$\mathbb{P}(S^n_2)\sim SO(n)\otimes_{\mathbb{R}}\mathbb{C}$

where $S^n_2$ is the space of homogeneous polynomials with degree 2 in $n$ variables and the isomorphism is as complex manifolds.

After discussing with a couple of people, I found that there were reasons to believe that this is true, since some topological properties of both spaces matched, but so far, I haven't found yet a formal proof of this fact.

In the case of this being true, it would turn out to be a really interesting property, since the object on the right is a Lie group, being isomorphic will imply that one can define a group structure in the set of conics, which is rather an interesting fact.

Saturday, May 22, 2010

Last Christmas, I went with my family to a road trip. I drove from Waco to New Orleans, a 8 hrs drive more or less, and when we reached there, I was a little tired. I laid down in the bed and stared at the roof of our room, and then, I started imaging a rubber ball bouncing all over a vertical cross section of our room, like a billiards ball, and then I wondered which kind of orbits could be ball have, depending of the incident angle of the first bounce.

After thinking a little, one can realize that for a rectangular cross section (my room's) this problem is not really hard, and the possible answers are quite few, depending on the ratio of the lengths of the rectangle. Then I though that a more interesting question would be when the cross section its a circle.

Basically two thing can happen, one is that the incident angle is such that the ball bounces only in finitely many places, and hence, the bouncing points make a periodic sequence on the circumference, and the other is that the bouncing points are dense in the circumference.

For instance, if the initial angle happens to be $\pi/2$, we are going to have only 2 bouncing points. Suppose that we have an initial angle $a$ and we are working with a unit circle, so that the arc length is the same as the angle value. If we look look up for periodic points, we require that $na=m\pi$, for some $n,m\in\mathbb{Z}^+$. That means that $a= \frac{m}{n}\pi$, and with out loss of generality, we can require $m/n \leq 1$. Then, one can draw the path of the ball as follows:

Step 1: Start drawing the path at vertex $k=1$

Step 2: Draw a line from the vertex $k$ to the vertex $k+m$ (sums are taken modulo $n$)

Step 3: Go to step 2 until you hit vertex 1

For instance, in the case $n=5$, we will have two different patterns

The orange path is obtained for $m=1$ and $m=4$, and the black one, for $m=2$ and $m=3$. In general, $m$ and $n-m$ will lead to the same orbit, but made in different directions (clockwise and counterclockwise).

So in the remaining case, when $a$ is not a rational multiple of $\pi$, we have that the bouncing points are dense in the circumference.

This result is not surprising, as this picture is related with the problem of wrapping a line around a torus, and it is well know that this wrapping is dense if it has an irrational slope, and periodic otherwise.

When glazed in this perspective, it turns out that the original problem, with the rectangle, and the one with the circle are exactly the same, since a torus is just the complex plane modulo a rectangle ( a lattice ), so at the end, my more interesting problem turned out to be as simple as the original one. It is quite interesting how apparently different problems became just two perspectives of the same phenomenon, one could seem totally boring and the other one, completely attractive and challenging. Some would say, beauty is in the eye of the beholder, but sometimes what happens is that differences are in the eye of the beholder.

Monday, March 29, 2010

A couple of weeks ago, I was reading some lecture notes on game theory and I came across a really neat game.

After discussing the very basics of game theory and decision making theory, the author of the lectures gives an exercise which I found really interesting and enjoyable, at the point that I went ahead and gave it as a quiz to my business calculus class.

To my surprise, most of my class got the right answer, which was truly a grateful feeling. The game is really simple so anybody can understand it, but in my opinion, it represents many aspects of real life.

It is as follows:

Every student is to write down a real $x_i$ number in between 0 and 10 inclusively. After doing so, one computes the mean $\bar{x}$ of all of the students' bets and each student's grade is given by

$10-\left|x_i-\frac{2}{3}\bar{x}\right|$

This might look a really simple task, and by no means a game at all, but it is a game of strategy and common sense.

Our desire as a students is to maximize our grade, but that depends on the average choice of the class, which might complicate a bit the analysis of a best strategy to pick our $x_i$.

It is not hard to see that a global best strategy is to pick $x_i=0$, as if everybody is a good and logical player, having all bets equal to $0$ would give each student's grade to be $10$, which is the best possible.

So, our personal best strategy should be to pick $0$, but in real life, not all players are good thinkers or really logical, so at the end of the day, our best strategy won't give us the best out come possible.

In a sense, we can think of this game as rewarding you if you somehow think average, and most of the times, the average thinking is not precisely the most wise and logical.

By the way the game was set up, we can see that it neither rewards the average thinking as much as someone that was 2/3 away from it. If you are 2/3 away from the average, you'll get full credit, and this is somehow what happens in real life. Usually not the average people get the best outcome nor the people that plays the best, but people that are in between.

This gives a really good example that in most occasions, your outcome does not depend only on your own strategy, but also in someone else's strategy, and that making the best decisions and taking the best choices does not guarantee your success.

Tuesday, February 16, 2010

Last week, in our quantum Mechanics class, we were going over symplectic spaces and symplectic transformations. A symplectic space is just a manifold together with a skew-symmetric non-degenerate bilinear form $J$ defined on it, and a symplectic transformation $S$ is a transformation of the manifold into itself such that it preserves $J$. One of the most common examples is when we take our manifold to be $M=\mathbb{F}^{2n}$ and the symplectic form

$J=\begin{pmatrix}0&I_n\\-I_n&0\end{pmatrix}$

where $F$ is a field and $I_n$ is the identity matrix. This is a symplectic manifold, and the set of symplectic transformations is know as $Sp(n,F)$. This is a well known Lie Group acting by multiplication on $M$, and one of its goodness is that this action is transitive, that is, for any non-zero $x,y\in M$, there is $S\in Sp(n,F)$ such that $y=Sx$.

This statement was actually part of our homework, to prove that the action is transitive, and I wanted to find a nicer way to prove it and not to do a proof that I had seen before in my previous courses, so I started thinking a bit of many different ways of saying that this action was transitive.

One way of seen this is by turning around the problem saying what would happen if we let $S$ to run over $Sp(n,F)$ and look at $Sx$ for $x$ fixed? Well, that is saying something like the orbit of $x$ is $M/\{0\}$ and that started to sound a bit familiar.

I was trying then to use some kind of orbit-stabilizer theorem and then use some cardinality argument and kill the problem. Although, I only did remember the finite version of this powerful theorem, which obviously, wouldn't help me at all, but in essence, that was what I was looking for. A cardinality argument would not help me in this situation, because I could have some proper subspace of the same cardinality of $M$ and this wouldn't lead me to the conclusion I was going after. Instead, a dimensionality argument was needed.

While searching for this and thinking what actually was going on behind the scenes in this group action, I saw how helpful is the notion of representation for understanding a strange object.

If $G$ is a Lie Group, we call a representation of it, a vector space $V$ in which $G$ acts on. We can think as $G$ be some sort of subgroup of $Gl(V)$, the set of linear transformations of $V$ into itself. For an element $x$ of $V$, we can talk about the $G$ orbit through $x$, $O_x$ as the set of all $g.x$ for $g\in G$. In some sense, $O_x$ is a copy of the shape of $G$. Also, from the geometrical point of view, a Lie Group is a manifold, endowed with superpowers (group structure) and hence, we can think of these orbits into $V$ as coordinate maps of $G$ given by $\phi(g)=g.x$, so really $O_x$ is how $G$ looks locally.

For example, take $O(2)$, which is the group of all $2\times 2$ matrices $O$ such that $OO^T=I$. This group is quite odd to picture, since it is a 1 dimensional manifold living in a 4 dimensional space, but by means of orbits, one can have a pretty good idea of how this group looks like. By picking a nonzero vector $x$ and looking at its orbit in $\mathbb{R}^2$, one can find that $O(2)$ looks locally like a circle.

In the general case, one can think as $G$ being a covering for $O_x$ and the degree of the cover is the number of connected components of $G$, for instance, in the above case, $O(2)$ has 2 connected components, the set of matrices with determinant equal to 1 and those of determinant equal to -1, and that fact is reflected in $O_x$ as the vector $g.x$ rotates counter clockwise for $O(2)_e$ (the identity component) and rotates clockwise in the other component, so each circle is drawn twice, and that means that $O(2)$ is a 2-fold cover for each $O_x$.

In this language, we can say that the stabilizer $G_X$ of an element $x$, is the fiber $\phi^{-1}(x)$ whose cardinality gives us the degree of the covering map.

Actually, from this point of view, $\phi$ defines a quotient map, which is very suitable for an orbit-stabilizer type argument. Since the stabilizer $G_x$ is a normal subgroup, one can think of $G$ as a principal $G_x$-bundle as $G/G_x\times G_x$ and making the identification $G/G_x\sim O_x$ and $G_x\sim \phi^{-1}(x)$ we have that $G\sim O_x\times\phi^{-1}(x)$.

Going away from counting arguments and going more into dimensionality, I found the so called Orbit-Stabilizer Theorem for Lie Groups which have the same feeling as the covering map approach. It states that

$dim(G)=dim(O_x)+dim(G_s)$

where $dim$ is regarded as manifolds.

In the $O(2)$ case, we have that $dim(G)=1$, $dim(O_x)=1$ and $dim(G_x)=0$ as any of the other cases when $G_x$ is a finite group, and hence, we have that $\phi$ is a quotient map and $dim(G)=dim(O_x)$ as expected from a covering map.

At the end, I didn't use any of these arguments for my proof, but I found quite enjoyable doing this diversion from my first thought.

Monday, January 25, 2010

This semester we started a course in Time Scales, which is an interesting generalization of the classic differential analysis. The idea of time scales is to provide a connection between the study of differential equations made on $\mathbb{R}$ and the study of difference equations on $\mathbb{Z}$.

This connection is made by taking a closed subset of $\mathbb{R}$ and start defining on it notions of a right and left derivatives, which are called the $\Delta$ and $\nabla$ derivatives.

On last week's class, one of my friends was talking about this definitions and also how would one define an integral using this time scales approach. As an example, he state the following integral

$\int_0^\infty e^{-\tau^2}\Delta \tau$

On the time scale $\mathbb{T}=\overline{\{q^\mathbb{Z}\}}$ for some $q>1$. This is just the closure of the set of all integer powers of $q$.

After dealing with the boring algebra involved, the previous integral has a value of

$(q-1)\sum_{n=1}^\infty (q^n+q^{-n})e^{-n^2}$

Finding the actual value of this expression boils down to calculate the value of

$\sum_{n=1}^\infty e^{an-n^2}$

This can be done by using the Jacobi Theta Function, which is given by

$\vartheta(z,w)=\sum_{n=-\infty}^\infty \exp(2\pi i nz+\pi i n^2 w)$

for $z\in\mathbb{C}$ and $w\in\mathbb{H}$. Thus letting $z=\frac{-ai}{2\pi}$ and $w=\frac{i}{\pi}$ gives the value that we are looking for. In this case, we have that

This is a really simple problem at first sight, but it caught my attention the fact that it all relies on determining the value for an expression that looks like

$\sum_{n=1}^\infty e^{p(n)}$

where $p(n)$ was a quadratic polynomial with a negative leading coefficient. A natural question came then to my mind, what would happen if we would have any polynomial instead?

My first idea was to study the case when $p(n)=-n^m$ for a fixed power $m$. The cases when $m=1,2$ are contained in the previous approach using the Jacobi Theta Function. I first try to calculate the series for some values of $m$, and I found that the above expression, as a function of $m$, converged really fast as $m$ was getting bigger.

Actually, it is not difficult to find out that this limit exists and has the value of

$\lim_{m\to\infty} \sum_{n=1}^\infty e^{-n^m}=\frac{1}{e}$

Following the same line, one can show that for $p(n)=-an^m$, with $a$ a constant, we have that

$\lim_{m\to\infty} \sum_{n=1}^\infty e^{-an^m}=\frac{1}{e^a}$

Thus, one could say that for a polynomial $p(n)=-an^m+O(n^{m-1})$ a good approximation its given by

$\sum_{n=1}^\infty e^{p(n)} \approx\frac{1}{e^a}$

Looking for a different approach towards a more precise answer, one can define a function given by $f(t)=\sum_{n=1}^\infty e^{p(n) t}$ and this function can be viewed as the heat kernel of a differential operator $P$ whose spectrum is given by $\sigma(P)=\{\lambda_n\}$, where $\lambda_n=p(n)$.

This could suggest the difficulty of such a closed form for $f(1)$, for instance, in the case of $p(n)=-n^m$, this would be related to the existence of an operator $P$ with eigenvalues $\{1/n\}$, which is one of the consequences of the Riemann Hypothesis.

Monday, January 4, 2010

After a little break time, I decided that a good way to start the year is by posting something that I was thinking on few days ago.

At the beginning of the Winter break, one of my friends back home post me a question about determining the foci a hyperbola just with compass and straightedge constructions. Thinking a little about it, I was trying to find some basic property of the foci of a hyperbola, and it came to my mind that they must have the same property no matter what kind of conic we are looking at, since all conics are equivalent under the $PSL(2,\mathbb{R})$ group.

So I tried to find some kind of property that the foci of all conics share, and one of them can be regarded as some kind of reflection property.

If you have a set of lines that passes through one of the foci of a conic and you take the reflection of these lines on the conic, the resulting set of lines passes through the other focus.

In the case of a circle, the two foci coincide in the center of the circle, so the condition holds, but this made me think of analyzing the orbit of a line in a circle. By this, what I mean is the following: start with a line that intersects the circle, then at the intersection points, reflect the line through the circle and keep with this process. Call the resulting intersection points on the circle, the orbit of the line. Now, a very natural question would be for which kind of lines does this orbit is finite? has a limit point? is dense in the circumference?

At the beginning, the answer seems really straightforward, one could say that the lines that lead to finite orbits are the ones that belong to sides of a regular polygon, which means that the lines leading to finite orbits are the ones whose distance to the center of the circle equals the apothem of some regular polygon inscribed in the circle.

However, this condition is quite weak, since there are more such lines. To look at these, we can take a different approach. One easy way is to look this is by taking the arc length instead. If we take the length of the circumference to be 1, then if the arc lengths of the pieces into which is divided the circle by the line are rationals, we have that the orbit is finite. Moreover, if the arc length is $q/p$, the orbit has length $p$.

This is equivalent to studying the circle regarded as $\mathbb{R}/\mathbb{Z}$, and here the orbit for a value $x$ is the set $\{[nx]: n\in\mathbb{Z}\}$ from where we have that it is finite iff $x\in\mathbb{Q}$ and it is dense in the circle otherwise.

Another interesting question would be to find sufficient conditions for finite orbits on a general conic, and this might be studied using the group structure of the circle $S^1$ and the projective properties of $PSL(2,\mathbb{R})$.